STRUCTURE OF BORON ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis and using the charged quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Boron (B) naturally occurs in two isotopes, B-10 and B-11, the latter of which makes up about 80% of natural boron. 14 radioisotopes have been discovered, with mass numbers from 6 to 21, all with short half-lives, the longest being that of B-8, with a half-life of only 770 milliseconds (ms) and B-12 with a half-life of 20.2 ms. All other isotopes have half-lives shorter than 17.35 ms, with the least stable isotope being B-7, with a half-life of 150 yoctoseconds (ys). Those isotopes with mass below 10 decay into helium (via short-lived isotopes of beryllium for B-7 and B-9) while those with mass above 11 mostly become carbon. ' ' WHY B-10 WITH S = +3 AND B-11WITH S = -3/2 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per neutron or per proton. In the case of B-10 using the diagram of B-10 we see that the protons at the corners of the rectangle make three bonds per proton which overcome the pp and nn repulsions. On the other hand the single bonds of p4 and p5 are stable ones because here the parallel spin along the z axis gives magnetic attraction like the magnetic attraction between the poles of two magnets. Therefore such pp systems contribute to the increase of the binding energy of bonds. Similarly the neutrons n4 and n5 give stable np bonds , because they have the same parallel spin along the z axis. Moreover in the diagram of the stable B-11 with S=-3/2 we observe the same stable structure as that of Be-9 in which the central neutron like the n3 makes with protons 4 bonds per neutron which contribute to the increase of the binding energy of bonds. Whereas in the diagram of the unstable B-9 you see that the protons make at the corners of the rectangle two bonds per proton unable to overcome the pp repulsions of long range. DIAGRAM OF Boron-9 with S = -3/2 Diagram of Boron-10 with S = +3 ' p4 (-1/2).n4 (- 1/2).p5(-1/2) n5(+1/2)p3(+1/2).n3(+1/2)p5(+1/2) ' ' n3(+1/2).p3(+1/2).n2(+1/2) n2(-1/2)..p2( -1/2) ' ' p1(-1/2). n1(-1/2).p2(-1/2) n4(+1/2)p1(+(1/2).n1(+1/2)p4(+1/2)' ' ' In the diagram of B-9 you see that there are 6 nucleons of negative spins and 3 nucleons of positive spins giving a total S = -3/2. In the following diagram of B-11 you see that we get the structure of the stable B-11 by adding in the structure of the stable Be-9 the p5 n6 system of strong vertical bond. Here we observe 7 nucleons of negative spins and 4 nucleons of negative spins giving S=-3/2. Whereas in the diagram of B-10 we observe 8 nucleons of positive spins and 2 nucleons of negative spins giving S =+3. '' ''DIAGRAM OF B-11 ' n4 (-1/2)..p4( -1/2)..n5(-1/2) ' ' p2(+1/2)..n3 (+1/2).p3(+1/2). n6(+1/2) ' ' ''' n1( -1/2)..p1(-1/2)..n2( -1/2)..p5(-1/2) ' '''STRUCTURE OF B-7 WITH S = -3/2 AND B-8 WITH S= +2 Adding the deuteron of S =-1 like the n2(-1/2)p5(-1/2) in the structure of Be-5 we get the structure of B-7 as shown in the following diagram of B-7. (See my STRUCTURE OF BERYLLIUM ISOTOPES ). ' DIAGRAM OF THE UNSTABLE B-7 WITH S = -3/2' Here you see 5 nucleons of negative spins and 2 protons of positive spins giving the total spin S =-3/2. In the absence of neutrons (for making a great number of pn bonds ) the pp repulsions of long range lead to the decay. ' p1(+1/2' ' p2(-1/2)..n1((-1/2)..p4(-1/2)..n2(-1/2)..p5(-1/2)' ' p3(+1/2)' On the other hand in the absence of n4 and n5 of the stable structure of B-10 we get the following diagram of the unstable B-8 with S = +2 in which the p1 and p3 make only 2 bonds per proton unable to overcome the pp repulsions of long range. In this unstable structure you see 6 nucleons of positive spins and 2 nucleons of negative spins giving the total S = +2. ' ' ' DIAGRAM OF B-8 WITH S =+2 ' ' p3(+1/2).n3(+1/2)p5(+1/2) ' ' n2(-1/2).p2( -1/2) ' ' p1(+(1/2).n1(+1/2)p4(+1/2)' STRUCTURE OF B-13, B-15, B-17, B-19 WITH S =-3/2 The structure of all these unstable nuclides is based on the structure of the stable B-11 with S =-3/2, in which extra neutrons of opposite spins make single bonds of weak horizontal binding energies leading to the beta decay. In the diagram of B-13 you see the two extra neutrons of opposite spins like the n7(+1/2) and the n8(-1/2) which lead to the beta decay. ' ' ' DIAGRAM OF THE UNSTABLE B-13 WITH S = -3/2' n4 (-1/2)..p4( -1/2)..n5(-1/2) ' ' n7(+1/2).......p2(+1/2)..n3 (+1/2).p3(+1/2). n6(+1/2) ' ' '' ''n1( -1/2)..p1(-1/2)..n2( -1/2)..p5(-1/2) .......n8(-1/2) Similarly in the structure of the unstable nuclide B-15 with S = -3/2, two more neutrons of opposite spins make single bonds, but they are not shown because the two weak horizontal bonds exist in front of protons and behind them. In other words B-15 has 4 extra neutrons of opposite spins giving the same S =-3/2 like that of B-11 with S =-3/2. In the same way we get the structure of B-17 with S = -3/2 and the structure of B-19 with the same S = -3/2 by adding extra neutrons of opposite spins. In both cases the extra horizontal bonds are not shown because they exist in front of protons and behind them. ' STRUCTURE OF B-12 WITH S = +1' In the following diagram of the B-12 you see that it has 7 nucleons of positive spins and 5 nucleons of negative spins giving a total S = 7(+1/2) +5(-1/2 ) = +1. Here the extra n7(-1/2) makes a single weak horizontal bond with p2(-1/2 which leads to the beta decay. ' DIAGRAM OF THE UNSTABLE B-12 WITH S = +1' ' n4 (+1/2)..p4( +1/2)..n5(+1/2)..p5(+1/2) ' ' n7(-1/2).....p2 ( -1/2)..n3 ( -1/2)..p3( -1/2)..n6( -1/2) ' ' ''' ''n1(+1/2)..p1(+1/2).. n2( +1/2) However, although the structure of B-16 is based on the structure of B-12 it gives S=0 because in the structure of B-16 there are 2 extra neutrons of negative spins which make two horizontal bonds with the p3(-1/2). Thus the total spin S=0 is given by S =+1 -1/2 -1/2 = 0 They are not shown in a diagram, because the horizontal bonds exist in front of p3 and behind it. Also B-16 has 2 extra neutrons of opposite spins which make horizontal bonds with p1(+1/2 and p2(-1/2). In other words the B-16 which has the structure of B-12 has two extra neutrons of negative spins giving S = -1 and two extra neutrons of opposite spins giving S=0. That is S = + 1 -1 + 0 = 0 STRUCTURE OF B-14 WITH S = -2 AND B-18 WITH S = -4 Using the diagram of B-13 with S = -3/2 we get the structure of B-14 with S = -2 by adding the extra neutron n9(-1/2) which makes a single horizontal bond with p1(-1/2). It is not shown in a diagram because it exists in front of the p1. Thus the total spin is given by S = -3/2 -1/2 = -4/2 = -2 However although the structure of B-18 is based on the structure of B-14 with S = -2 we see that here the spin is S =-4. It is due to the fact that the 4 extra neutrons like the n10(-1/2), the n11(-1/2), the n12(-1/2) and the n13(-1/2) make 4 single horizontal bonds with p4(-1/2) and p5(-1/2). Thy are not sown in a diagram because they exist in front of protons and behind them. Here the total spin is given by S = -2 -1/2 -1/2 -1/2 -1/2 = -4. Category:Fundamental physics concepts